Answer:
51.9
8.9
78.5
8.8
Explanation:
You round up if the number to the right is 5 or greater.
So, apply this to the numbers and the 8 in A rounds up, going to a 51.9.
The 9 does not round up because of the one.
The 5 also does not round up because of the 2.
The 7 does round up because of the 6.
Question 1.x - 7 > - 8
Adding 7 to both sides, we get:
x - 7 + 7 > - 8 + 7
x > -1
Thus the answer to the inequality is option Fourth.Question 2.
A number exceeds 66. Let that number be x. The number exceeds 66 means that the number is larger than 66. So in form of an expression we can write the inequality as:
x is greater than 66
x > 66
So, option 1st gives the correct answer.Question 3.The tiles are square shaped and area of a square can be calculated as the square of its Length.
Area of square = (Length)²
If we are given the Area, we can find the length as:
Length =

For Tile A, the length will be:
So length is a Rational number
For Tile B, the length will be:
So length is a Rational number
For Tile C, the length will be:
So, Length is not Rational.
For Tile D, the length will be:
Length is not Rational
Thus, the lengths of Tile A and Tile B are rational only.
Therefore, the correct answer is 1st option
(9/6) = (15/?)
9? = 90
? = 10
Answer:
bdbdjsjsbsbsbbshssbdjdjdjsjhshshdjdd
the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.
from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.
the standing up sides are simply rectangles of 8x3.
if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D5%5C%5C%20p%3D24%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%285%29%2824%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bjust%20for%20one%20octagon%7D%7D%7BA%3D60%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwo%20octagon%27s%20area%7D%7D%7B2%2860%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Beight%20rectangle%27s%20area%7D%7D%7B8%283%5Ccdot%208%29%7D%5Cimplies%20120%2B192%5Cimplies%20312)